2023 H2 Paper 1 Solutions Students

2023 A Level Paper 1 Suggested Solutions 1 Find the exact equation of the tangent to the curve ( ) 2 ln 11 5yx=− at the point where 2.x= [5] Solutions ( ) ( )( ) ( ) 2 ln 11 5 Differentiating with respect to : 1d 2 11 5 5 10 11 5 Equa d dAt 2, e, 10e d t 2 e 10e t ( 2) 10e 21e ion of the tangent a yx x y xxyx yxy x x yx yx =− = − − =− − = = =− = − =− − =− + 2 The first four terms of a sequence are given by 1u = 10, 2u = 61, 3u = 206 and 4u = 469. It is given that nu is a cubic polynomial. (a) Find nu in terms of n. [3] Solutions Since nu is a cubic polynomial, 32 nu an bn cn d= + + + for some a, b, c, d  ℝ. ( )1 10 1u a b c d= + + + = −−− ( )2 8 4 2 61 2u a b c d= + + + = −−− ( )3 27 9 3 206 3u a b c d= + + + = −−− ( )4 64 16 4 469 4u a b c d= + + + = −−− Using GC to solve (1), (2), (3) and (4), a = 4, b = 23, c = – 46, d = 29 324 23 46 29nu n n n = + − +

2023 A Level Paper 1 Suggested Solutions (b) Find the range of values of n for which nu is greater than 25 000. [2] Solutions 324 23 46 29nu n n n= + − + > 25 000 From GC, 16.9n i.e. 17n 3 Vectors a and b are such that ab = –1. It is also given that ( ab + a) is perpendicular to (ab + b). (a) Show that |ab| = 1. [3] Solutions Given that ( )a b a+ is perpendicular to ( )a b b+  ( ) ( ).0a b a a b b +  + = ( ) ( ) ( ) ( ) 2 . . . . 0 0 0 . 0 a b a b a b b a a b a b a b a b   +  +  + =  + + + = [since ab is perpendicular to both a and b ] 2 2 0 0 1 0 1 1 since 0 (shown) ab ab a b a b  + + − = =  =   y y = 25000 y = 4x3 + 23x2 – 46x + 29 16.86920 x

2023 A Level Paper 1 Suggested Solutions (b) Hence find the angle between the direction of a and the direction of b. [3] Solutions Let the angle between the direction of a and the direction of b be  . ( ) ( ) ( ) ( ) sin 1 1 cos 1 2 1 : tan 12 a b a b a b a b     = = −−−  = =− −−− =− Since tan 0 is obtuse and lies in the second quadrant. 1180 tan 1 135 − = − − =  4 (a) Find cos cos dpx qx x , where p and q are constants such that p  q and p  – q. [2] Solutions ( ) ( ) ( ) ( ) cos cos d 1 cos cos d2 sin sin1 ,2 where is an arbitrary constant px qx x p q x p q x x p q x p q x Cp q p q C = + + −  + − = + + +−   (b) Given that n  0, show that 2 sin coscos d x nx nxx nx x c nn= + + , where c is an arbitrary constant. [3] Solutions 2 sin sincos d d sin cos (shown) where is an arbitrary constant nx nxx nx x x x nn x nx nx cnn c =− = + + 

2023 A Level Paper 1 Suggested Solutions (c) Using the result in part (b) show that, for all positive integers n, the value of π 0 cos dx nx x can be expressed as 2 k n , where the possible value(s) of k are to be determined. [2] Solutions   ( ) ( ) π π 20 0 22 22 22 2 sin coscos d πsin π cos π cos00 cos π10 sin π 0 for all positive integers 1 1 cos π 1 for all positive integers 2 when isodd 0 when n n x nx nxx nx x nn nn n n n n nn nn nn nn nn =+    = + − +       = + −  = −=−  =− − =  isevenn   0k= or 2k=− (d) Using the result in part (b) find the exact value of π 0 cos2 dx x x . [3] Solutions From the graph (above),

2023 A Level Paper 1 Suggested Solutions πcos2 , 0 4cos2 ππcos2 , 42 x x x xx x x x  = −   π 2 0 ππ 42 π0 4 ππ 42 π0 4 cos 2 d cos 2 d cos 2 d sin 2 cos 2 sin 2 cos 2 2 4 2 4 π 1 1 π0 0 0 08 4 4 8 π 4 x x x x x x x x x x x x x x x = + −    = + − +          = + − − − − − −       =   5 (a) Use the method of difference to show that ( )( ) 2 2 11ln n r rr r=  − +     = 1ln ln 2n n + − . [3] Solutions ( )( ) 2 2 11ln n r rr r=  − +     ( ) ( ) 2 ln 1 2ln ln 1 n r r r r = =  − − + +  ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ln1 2ln 2 ln 3 ln 2 2ln 3 ln 4 ln 3 2ln 4 ln 5 ... ln 3 2ln 2 ln 1 ln 2 2ln 1 ln ln 1 2ln ln 1 n n n n n n n n n  −+ + − + + − + =+ + − − − + − + − − − + + − − + + = 0 – ln 2 – ln n + ln (n + 1) 1ln ln 2n n +=−  (Shown) (b) Show that the corresponding infinite series is convergent and state the sum to infinity. [2] Solutions ( )( ) 2 2 11 1ln ln ln 2 1ln 1 ln 2 n r rr n rn n =  − +  +=−  = + − 

2023 A Level Paper 1 Suggested Solutions As n→ , 111 n+→ , ( )( ) 2 2 11ln ln1 ln 2= ln 2 n r rr r=  − +  → − −   , a finite unique value. Therefore, the series is convergent and the sum to infinity = –ln 2 (c) Show that ( )( )20 2 10 11ln ln r rr a rb=  − +  =   , where a and b are integers to be found. [2] Solutions ( )( ) ( )( ) ( )( ) 20 2 10 20 9 22 22 11ln 1 1 1 1ln ln 21 10ln ln 2 ln ln 220 9 21 9ln 20 10 189ln 200 r rr rr r r r r r rr = ==  − +     − +   − + =−               = − − −             =  =   Therefore, a = 189 and b = 200 6 (a) Using double angle formulae, show ( ) 4 1cos cos4 4cos2 38  = + + . [2] Solutions ( ) ( ) ( ) 22 4 2 2 2 cos cos 1 cos 2 1 , since cos 22 1 cos 2 2cos 2 14 11 cos 4 1 2 2cos cos 2 142 1 cos 4 4cos 2 3 (shown)8 1         =− = =+ = + + = + + + = + + Alternative

2023 A Level Paper 1 Suggested Solutions ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 22 4 4 11cos 4 4cos 2 3 2cos 2 1 4cos 2 388 2 cos 2 2cos 2 18 1 cos 2 14 1 2cos 1 14 1 4cos4 cos shown          + + = − + + = + + =+ = − + = = (b) The region R lies in the first quadrant and is bounded by the curve ( ) 342 9yx=− , the x- axis and the lines x = 1.5 and x = 3. R is rotated about the x-axis through 2 radians. Using the substitution x = 3sin, find the exact volume generated. [6] Solutions x = 3sin d 3cosd x = 1.5 sin 0.5 6 3 sin 1 2 x x   =  =  = =  =  = Required volume 3 2 1.5 =πd yx ( ) ( ) ( ) ( ) 33 2 2 1.5 π 3 22 2 π 6 π 3 22 2 π 6 =π 9 d dπ 9 9sin d d π 9 9sin 3cos d xx x     − =−  =−   

2023 A Level Paper 1 Suggested Solutions ( ) ( ) ( ) ( ) π 3 22 2 π 6 π 32 π 6 π 42 π 6 π 2 π 6 3 π 2 π 6 81π 1 sin cos d 81π cos cos d =81π cos d 181π cos 4 4cos 2 3 d8 81 sin 4π 2sin 2 384 81 3 π 3 ππ 0 0 38 2 8 2 81 9 3π π units88             =− = = + + = + + = + + − − −  =−      

2023 A Level Paper 1 Suggested Solutions 7 The function f is defined by 24f : , , 3. 3 xx x x x +→  − (a) Sketch the graph of f ( ),yx= giving the equations of any asymptotes and the coordinates of the points where the curve meets the axes. [3] Solutions 2 4 10 233 xy xx += = − +−− Equations of asymptotes: 22y= − = , 3x= When 0x= , 44 33y== When 0y= , 2x=− (b) Hence state the range of f. [1] Solutions  )fR 0,= (c) Explain why the function 2f does not exist. [1] Solutions  ) ( ) ( )ffR 0, , D ,3 3,=  = −   Since f3R but f3D ff 2 Hence R D f does not exist.   The domain of f is further restricted to ,xa where a is a constant. (d) State the greatest value of a such that the function 1f − exists. [1] y x = 3

2023 A Level Paper 1 Suggested Solutions Solutions Greatest value of 2a=− (e) Hence find 1f ( )x− and state its domain. [4] Solutions ( ) ( ) ( ) ( ) ( ) 1 1 f 1 f When 2, 2 4 2 4f 33 Let f 3 2 4 2 4 3 43 2 Since f , 43f, 2 D R [0,2) x xxx xx yx y xy x x y y yx y xy xx x − − − − ++= =−−− = − + = + − = + += − = += − == 8 Do not use a calculator in answering this question. (a) (i) Express z, where 1 3i,z=− + in the form ie,r  where 0r and .  −   [2] Solutions ( ) ( ) ( ) 22 1 2i π3 1 3i 1 3 2 32arg π tan π 13 2e z z z z − =− + = − + = = − =  = (ii) Find the smallest posi